Do the premises logically imply the conclusion? $$b\rightarrow a,\lnot c\rightarrow\lnot a\models\lnot(b\land \lnot c)$$
I have generated an 8 row truth table, separating it into $b\rightarrow a$, $\lnot c\rightarrow\lnot a$ and $\lnot (b\land\lnot c)$. I know that if it was 
$$\lnot c\rightarrow\lnot a\models\lnot(a \land \lnot c)$$
I would only need to check the right side with  every value that makes the left side true to make sure the overall statement is true. How do I deal with more than one premise?
 A: The second premise $\neg c\to\neg a$ implies that $a\to c$.
The first premise $b\to a$ leads to $b\to a\to c$, which implies $\neg c\to\neg b$.
The two last statements clearly prevent $b\land\neg c$ from being true.
A: So you can show this a couple of ways. One of them is to use truth table semantics- your hypotheses are connected by an "and". Another equivalent, but faster way is to use the method of analytic tableaux, which essentially works by trying to find ways to falsify the statement (if they all work out to contradictions, then it must be true).
A third method you can use is called sequent calculus. This works in Intuitionistic Propositional Calculus, which does not use the law of excluded middle. The cool thing about IPC is that using sequent calculus can actually produce functions that verify the formula.
$b \Rightarrow a, \lnot c \Rightarrow \lnot a \vdash \lnot (b \land \lnot c) \;\;\; \mathrm{notR}$
$\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, (b \land \lnot c) \vdash false \;\;\; \mathrm{andL}$
$\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b,\lnot c \vdash false \;\;\; \mathrm{impL}$
$\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b,\lnot c \vdash b \;\;\; \mathrm{axiom}$
$\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a \vdash false \;\;\; \mathrm{impL}$
$\;\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a \vdash \lnot c \;\;\; \mathrm{axiom}$
$\;\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a, \lnot a \vdash false \;\;\; \mathrm{notL}$
$\;\;\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a, \lnot a \vdash a \;\;\; \mathrm{axiom}$
The evidence that this produces is the following function (unless I made a mistake in forming the evidence term): $\lambda f. \lambda x. (f_1(x_1),f_2(x_2))$
